Series and sequences #

What is a series #

To evaluate series, first find the partial sum:

Find the formula for \(S_n\)

$$ S_n= \frac{n(n+1)}{2} $$

Take the limit as \(n \rightarrow \infty\)

$$ \lim_{n \rightarrow \infty} \frac{n(n+1)}{2} = \infty $$

Telescoping #

These series look like two repeating fractions that end up canceling everything except something from the first term and something from the last. For example:

First find the partial sum \(S_n\)

Almost all of these fractions will cancel if you see the patern. The only 2 left are:

$$ S_n = - \frac{1}{3} + \frac{1}{2n+3} $$

Take the limit of this partial sum \(S_n\)

$$ \lim_{n \rightarrow \infty}\big[ - \frac{1}{3} + \frac{1}{2n+3} \big] = - \frac{1}{3} $$

Geometric #

Geometric series take the form of:

$$\sum_{n=1}^\infty ar^{n-1}$$

The series will converge of \(\big|r\big|< 1\) , otherwise it will diverge.

If the sum does converge, the sum is:

$$\sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r}$$

Shortcut #

If the first power of the sequence is 0 then the first term is \(a\) . \(a\) stands for the first term in your series.

For example:

Another example:

The mistake most people make here is thinking that \(a = \frac{1}{3}\) . This isn’t the case because plugging in \(n=2\) doesn’t make the first exponent 0. So split off more $\frac{e}{3}$’s to make it in the right form:

Now since the first term makes the exponent go to 0. You can tell what \(a\) and \(r\) are now. So:

$$ = \frac{ \frac{e^2}{27} }{1- \frac{e^2}{3}} $$

So the shortcut here is that you can start with

$$ = \sum_{n=2}^{\infty} \frac{1}{3} \Big( \frac{e}{3}\Big)^n $$

and simply plug in 2 for \(n\) (the starting point). Since we know that the first term is \(a\) you can jump to the answer:

$$ = \frac{ \frac{e^2}{27} }{1- \frac{e^2}{3}} $$

Harmonic #

Harmonc series are defined as:

$$ \sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} +\ …\ + \frac{1}{n} $$

Harmonic series are divergent. If a sequence \(\{a_n\}\) is convergent, any subsequence of \(\{a_n\}\) must also be convergent. To show that a sequence \(\{a_n\}\) diverges, it is enough to show that a subsequence diverges.

Note: If a series converges, then

So to show a series diverges, it’s enough to show:

$$ \lim_{n \rightarrow \infty} a_n \neq 0 $$

If the limit doesn’t equal 0, or \(DNE\) , the series \(\{a_n\}\) diverges.

Remember! The limit equaling 0 does NOT necessarily mean convergence!

Example #

Diverges because the limit equals 0!

Properties of convergent series #

1. You can always pull a constant out in front of the series

$$ \sum_{n=1}^{\infty} C a_n = C \sum_{n=1}^{\infty} a_n $$

2. You can split sequences on sums or differences

$$ \sum_{n=1}^{\infty} (a_n \pm b_n) = \sum_{n=1}^{\infty} a_n \pm \sum_{n=1}^{\infty} b_n $$

Example #

$$ \begin{split} \sum_{n=1}^{\infty} \Big ( \frac{2^n-5^n}{3^n}\Big ) &) \ & =\sum_{n=1}^{\infty} \frac{2^n}{3^n} - \sum_{n=1}^{\infty} \frac{5^n}{3^n} \end{split} $$

If we split this sequence into parts, each part much be convergent for the entire sequence to be convergent! If any single part is divergent then the entire thing is divergent.

$$ =\sum_{n=1}^{\infty} \Big (\frac{2}{3}\Big )^n - \sum_{n=1}^{\infty} \Big (\frac{5}{3}\Big )^n $$

In this form we can evaluate them as geometric series. Automatically we know this is divergent because the rightmost fraction’s \(r\) is greater than 1. Since a subsequence of the original diverges, the original does too.

P series #

$$ \sum_{n=1}^\infty \frac{1}{n^P} $$

When \(P < 1\) the series will diverge.

$$ \lim_{n \rightarrow \infty} \frac{1}{n^P} $$

When \(P > 1\) the series will converge (can be shown with the integral test).

Example 1 #

$$ \sum_{n=1}^{\infty} \frac{1}{n^2} $$

Here we can see that \(P\) , so the series must converge.

Example 2 #

$$ \sum_{n=1}^\infty \frac{1}{\sqrt[3]{n}} $$

Here we can see that \(P\) , so the series diverges.

Example 3 #

$$ \sum_{n=1}^\infty n^{-\pi} = \sum_{n=1}^\infty \frac{1}{n^\pi} $$

Since \(P\) the series must converge.